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Update: prevent superfluous set updates #18

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oliverfoster merged 2 commits into from
Jul 10, 2025
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Update: prevent superfluous set updates #18

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1c29a29
Update: prevent superfluous set updates (fixes #17).
danielghost Jun 27, 2025
120d455
Remove duplicate sets from intersection lookups.
danielghost Jun 27, 2025
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5 changes: 3 additions & 2 deletions js/adapt-contrib-scoring.js
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Original file line number Diff line number Diff line change
Expand Up @@ -83,9 +83,10 @@ class Scoring extends Backbone.Controller {
* @fires Adapt#scoring:update
*/
update() {
const updateSubsets = !this._queuedChanges?.length
const queuedChanges = [...new Set(this._queuedChanges)];
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@danielghost danielghost Jun 30, 2025

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Is it better to check if the queue already contains the model before adding, rather than removing duplicates later?

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@oliverfoster oliverfoster Jul 10, 2025 •
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It makes no difference with arrays at this size.

It is difficult to quantify which is more performant.

It may be cheaper to remove duplicates as the list grows, as at each addition, there would be fewer duplicate comparisons and the list would remain as short as possible. However, the overhead of performing a duplicate comparison each time may cost more than performing a single duplicate check on a larger set later.

It entirely depends on the quantity of duplicate and non-duplicate items, and the frequency of additions. Then, whether it is more important to add quickly and process slowly or vice-versa, or if RAM usage is important than processing speed. etc

const updateSubsets = !queuedChanges?.length
? this.subsets
: this._queuedChanges.reduce((updateSubsets, model) => updateSubsets.concat(getSubsetsByModelId(model?.get('_id'))), []);
: [...new Set(queuedChanges.reduce((subsets, model) => subsets.concat(getSubsetsByModelId(model?.get('_id'))), []))];
updateSubsets.forEach(set => set.update());
this._queuedChanges = [];
if (!updateSubsets.length) return;
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